I’ve been studying Spivak’s Calculus on Manifolds and I’m really not getting what’s behind partitions of unity. Spivak introduces the topic with the following theorem:

Let A⊂Rn and let O be an open cover of A. Then there is a collection Φ of C∞ functions φ defined in an open set containing A, with the following properties:

For each x∈A we have 0≤φ(x)≤1.

For each x∈A there is an open set V containing x such that all but finitely many φ∈Φ are 0 on V.

For each x∈A we have ∑φ∈Φφ(x)=1 (by 2 for each x their sum is finite in some open set containing x).

For each φ∈Φ there is an open set U in O such that φ=0 outside of some closed set contained in U.

The point is that I’ve heard that partitions of unity are able to transfer local results to global results, and this is of great importance, but I’m not really getting the intution behind this theorem. I mean,

whya collection of functions with these four properties is able to do such job?When I see a theorem/definition, I try to really get the intution behind it: “why we should really think about doing things this way”, because I think that this is a good way to understand what we are doing, but with partitions of unity I’m really not getting the idea.

While Spivak uses this just for integration on Calculus on Manifolds for what I’ve seem, in his Differential Geometry books he starts to use it really more generally to get global results from local ones (obtained with charts).

So, given the great importance of this topic, what’s the real intuition behind this theorem and partitions of unity in general?

**Answer**

In a few words, the point of partitions of unity is to take functions (or differential forms or vector fields or tensor fields, in general) that are locally defined, bump them off so they’re smoothly 0 outside their domain of definition, and then add them all up to get something globally defined.

For example, suppose you have a surface S in R3 that you can *locally* write as f=0, but you don’t know how to do so globally. You can cover S with open sets Ui⊂R3 on which you have smooth functions fi:Ui→R with S∩Ui={x∈Ui:fi(x)=0}. Consider Φ={ϕi}, where ϕi is supported in Ui. Then f=∑ϕifi will define a smooth function with f=0 on S. If you want f to be zero *only* on S, you can take an additional open sets U0=R3−S, set f0=1, and throw ϕ0f0 into your sum.

**Attribution***Source : Link , Question Author : Gold , Answer Author : Ted Shifrin*